Let f be the function f(x) = xu-ux, which maps K into itself.
Let C be Zp[u], a finite field in K. Multiply by C on the left, and K becomes a left C vector space.
Since C commutes with u, f(C) = 0. also, f is linear, and f(C*x) = C*f(x). Thus f is a linear map on K, when viewed as a left C vector space.
Let R be the ring of endomorphisms of K, when viewed as a left C vector space. Note that R, like K, has characteristic p.
Think of f as the difference between two endomorphisms, namely x*u and u*x. Within R, these two endomorphisms commute; either way we get u*x*u. Since they commute, one can expand fp by the binomial theorem. The result is (xu)p-(ux)p. Remember that fp does not mean f times f times f … p times, but rather, p applications of f in R. Similarly, (xu)p is the application of the function xu, p times, which is xup. Therefore fp = xup-upx.
As a finite field, let C have order pm. Invoke f pm times, and since u raised to the pm = u, the result is xu-ux, or f. Within R, f to the pm = f.
Write fpm-f = 0, the 0 endomorphism, and factor fpm-f in R. Let b be an element of C and consider the endomorphism bx. Verify that bx and f commute; you can apply them in either order. Expand the product of f-bx for every b in C. Set f-0 to the side. That leaves all the nonzero elements of C. These are the distinct roots of 1, and the solutions to the polynomial xpm-1-1. Thus the expansion of this product is fpm-1-1, and when we bring f-0 back in, the result is fpm-f. Therefore the composition of endomorphisms f-bx yields 0.
Since these functions commute, apply f first, then the other functions. Suppose all the other functions f-bx are injective. Their composition is injective, and the result cannot be 0 unless f is 0. Since u is not central, f cannot be 0. Therefore f-bx is not injective for some b.
Let x be a nonzero element in the kernel. For this particular x, write the following:
f(x) = bx
xu-ux = bx
xu = (u+b)x
xu/x = u+b
The conjugate of a nonzero element is nonzero, so u+b is a nonzero element in C. Since b is nonzero, u and u+b are distinct elements.
The order of u in K, whatever that may be, is the same as the order of xu/x in K. Therefore u and u+b are both elements of C exhibiting the same order. They both generate the same multiplicative cyclic subgroup in C*. Thus u+b is some power of u beyond u1. We have found an x such that xu/x = ui.
However, we are looking for a commutator. Set y = xu-ux. Since u and x don't commute, y is nonzero. Multiply xu = uix on the right by u, and on the left by u, and subtract the resulting equations. The result is yu = uiy. Divide by y on the right and yu/y = ui. That completes the proof.